# Thread: A messy Indefinite Integral

1. ## A messy Indefinite Integral

$\int \frac{1+u+u^2}{1 - u - u^2 - u^3} \, du$

While helping disclaimer with the differential equation problem, I ran into this integral. I am having a hard time with it. So please help.

Thanks,
Isomorphism

2. Originally Posted by Isomorphism
$\int \frac{1+u+u^2}{1 - u - u^2 - u^3} \, du$

While helping disclaimer with the differential equation problem, I ran into this integral. I am having a hard time with it. So please help.

Thanks,
Isomorphism
That's because it is a stinker to integrate. It looks like it's going to be done using a partial fraction expansion, but as the zeros of the denominator are complex I'm leaving it alone. (This is why I didn't respond to the other post.)

-Dan

3. Originally Posted by topsquark
That's because it is a stinker to integrate. It looks like it's going to be done using a partial fraction expansion, but as the zeros of the denominator are complex I'm leaving it alone. (This is why I didn't respond to the other post.)

-Dan
Ya I thought so. But does that mean there is a better substitution?

P.S: The online integrator gave me a scary answer

4. Hello,

I cheated a little and asked my TI...

There is a real root and 2 complex roots...

... and they look awful

5. Originally Posted by Peritus
That looks much neater than the wolfram online integrators solution. However they have just neatly hidden the ugly ones in that big-Oh notation, isnt it?

6. Hello,

Originally Posted by Plato
$\frac{{1 + u + u^2 }}
{{1 - u - u^2 + u^3 }} = \frac{3}
{{2\left( {u - 1} \right)^2 }} - \frac{3}
{{4\left( {u - 1} \right)}} + \frac{1}
{{4\left( {u + 1} \right)}}
$
But it's $1-u-u^2{\color{red}-}u^3$

7. The integral in question is $\int \frac{1+u+u^2}{1 - u - u^2 - u^3} \, du$

Originally Posted by Plato
$\frac{{1 + u + u^2 }}
{{1 - u - u^2 + u^3 }} = \frac{3}
{{2\left( {u - 1} \right)^2 }} - \frac{3}
{{4\left( {u - 1} \right)}} + \frac{1}
{{4\left( {u + 1} \right)}}
$
So how will that help? Is there some manipulation I cannot see?

8. If it was +u^3 it would have been easier

9. Originally Posted by Moo
If it was +u^3 it would have been easier
Yes I know... I wished the same while working on it. I even rechecked the computations to see if I messed up with some sign.... But in vain.....

10. I was going to comment on the fact that integrals shouldn't be classified as messy or nasty, but rather challenging.

Then I saw what you were talking about, and I must concur with the adjective you used.

EEEK!

11. There's no way to compute this, you can't even factorise the denominator as a "nice" form.